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1230. Toss Strange Coins πŸ”’

Description

You have some coins.  The i-th coin has a probability prob[i] of facing heads when tossed.

Return the probability that the number of coins facing heads equals target if you toss every coin exactly once.

 

Example 1:

Input: prob = [0.4], target = 1
Output: 0.40000

Example 2:

Input: prob = [0.5,0.5,0.5,0.5,0.5], target = 0
Output: 0.03125

 

Constraints:

  • 1 <= prob.length <= 1000
  • 0 <= prob[i] <= 1
  • 0 <= target <= prob.length
  • Answers will be accepted as correct if they are within 10^-5 of the correct answer.

Solutions

Solution 1: Dynamic Programming

Let $f[i][j]$ represent the probability of having $j$ coins facing up in the first $i$ coins, and initially $f[0][0]=1$. The answer is $f[n][target]$.

Consider $f[i][j]$, where $i \geq 1$. If the current coin is facing down, then $f[i][j] = (1 - p) \times f[i - 1][j]$; If the current coin is facing up and $j \gt 0$, then $f[i][j] = p \times f[i - 1][j - 1]$. Therefore, the state transition equation is:

$$ f[i][j] = \begin{cases} (1 - p) \times f[i - 1][j], & j = 0 \ (1 - p) \times f[i - 1][j] + p \times f[i - 1][j - 1], & j \gt 0 \end{cases} $$

where $p$ represents the probability of the $i$-th coin facing up.

We note that the state $f[i][j]$ is only related to $f[i - 1][j]$ and $f[i - 1][j - 1]$, so we can optimize the two-dimensional space into one-dimensional space.

The time complexity is $O(n \times target)$, and the space complexity is $O(target)$. Where $n$ is the number of coins.

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class Solution:
    def probabilityOfHeads(self, prob: List[float], target: int) -> float:
        n = len(prob)
        f = [[0] * (target + 1) for _ in range(n + 1)]
        f[0][0] = 1
        for i, p in enumerate(prob, 1):
            for j in range(min(i, target) + 1):
                f[i][j] = (1 - p) * f[i - 1][j]
                if j:
                    f[i][j] += p * f[i - 1][j - 1]
        return f[n][target]

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class Solution {
    public double probabilityOfHeads(double[] prob, int target) {
        int n = prob.length;
        double[][] f = new double[n + 1][target + 1];
        f[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j <= Math.min(i, target); ++j) {
                f[i][j] = (1 - prob[i - 1]) * f[i - 1][j];
                if (j > 0) {
                    f[i][j] += prob[i - 1] * f[i - 1][j - 1];
                }
            }
        }
        return f[n][target];
    }
}

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class Solution {
public:
    double probabilityOfHeads(vector<double>& prob, int target) {
        int n = prob.size();
        double f[n + 1][target + 1];
        memset(f, 0, sizeof(f));
        f[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j <= min(i, target); ++j) {
                f[i][j] = (1 - prob[i - 1]) * f[i - 1][j];
                if (j > 0) {
                    f[i][j] += prob[i - 1] * f[i - 1][j - 1];
                }
            }
        }
        return f[n][target];
    }
};

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func probabilityOfHeads(prob []float64, target int) float64 {
    n := len(prob)
    f := make([][]float64, n+1)
    for i := range f {
        f[i] = make([]float64, target+1)
    }
    f[0][0] = 1
    for i := 1; i <= n; i++ {
        for j := 0; j <= i && j <= target; j++ {
            f[i][j] = (1 - prob[i-1]) * f[i-1][j]
            if j > 0 {
                f[i][j] +=